The task is to find the absolute max and absolute min of this function:

$f(x,y) = 4xy^2-x^2y^2-xy^3$

on the domain D: $D={(x,y) | x\ge0, y\ge0, x + y \le 6}$

So I get the partial derivatives of $f(x,y)$ to find the critical points

$f_x(x,y) = 4y^2-2xy^2-y^3$

$f_y(x,y) = 8xy-2x^2y-3xy^2$

Solving these for zero I get the critical point of (0,0). and f(0,0) = 0.

Then I find the extreme values at points on domain D which are (0,0),(6,0),(0,6).

In all cases the extreme values f(0,6) and f(6,0) are zero. I get the feeling I'm doing something wrong here. What are the absolute min/max values?


Your absolute max and min will be achieved at critical points in the domain's interior, or on the boundary of the domain.

To find critical points (if any) in the interior of $D$, set $f_x(x,y)=y^2(4-2x-y)$ and $f_y(x,y)=xy(8-2x-3y)$ equal to $0$, and assume that $x>0,y>0,$ and $x+y<6$. Since $x,y\neq 0$ and $$y^2(4-2x-y)=0\\xy(8-2x-3y)=0,$$ it follows that $$4-2x-y=0\\8-2x-3y=0.$$ Solving this system for $y$ yields $y=2$, and back-substitution tells us that $x=1$. Indeed, $x+y=3<6$ and $x,y>0$, so this critical point is in the interior of $D$.

You've seen that $f(0,6)=f(6,0)=f(0,0)=0$, so it remains to check the boundary curves of $D$. On the lines $x=0$ and $y=0$, $f$ is constantly zero. All that's left is to check along the line $y=6-x$ for $0<x<6$. We do this by letting $$\begin{align}g(x) &:= f(x,6-x)\\ &= 4x(6-x)^2-x^2(6-x)^2-x(6-x)^3\\ &= (4x-x^2)(6-x)^2-(6x-x^2)(6-x)^2\\ &= -2x(6-x)^2.\end{align}$$ Then we check for critical points of $g$ by setting $g'(x)=0$ and looking for solutions with $0<x<6$. You should find such an $x$, and then you will also need to check $(x,6-x)$ as a potential location for the global max or min of $f$.

  • $\begingroup$ Great. Got f(2,4) = -64 as min and f(1,2)=4 as max. $\endgroup$ Mar 13 '13 at 22:03

You need to determine the extrema on the boundary, the extrema on the critical points of the boundary and the extrema of the interior.


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